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`K_(a)` for HCN is `5.0xx10^(-10)` at `25^(@)C`. For maintaining a constant pH of 9. Calculate the volume of `5.0M KCN` solution required to be added to 10 mL of `2.0M HCN` solution.A. 2 mLB. 5 mLC. 3mLD. 4mL |
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Answer» Correct Answer - A `pH=pK_(a)+log.(["Salt"])/(["Acid"])` `9=-log (5xx10^(-10))+log.(["Salt"])/(["Acid"])` or `log. (["Salt"])/(["Acid"])=9+log (5xx10^(-10))=1.6990` or `(["Salt"])/(["Acid"])=`antilog`(1.6990)=0.5` Ph `=-log K_(a)+log.(["Salt"])/(["Acid"])` |
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